Chapter 2. Sequences 1. Limits of Sequences

1 Chapter 2. Sequences 1. Limits of SequencesLetAbe a nonempty set. A function from IN toAis called asequenceof elementsinA. We often use (an)n=1;2;:::to denote a sequence. By this we mean that a functionffrom IN to some setAis given andf(n) =an Aforn IN. More generally, a functionfrom a subset ofZZtoAis also called a is important to distinguish between a sequence and its set of values. The sequence(an)n=1;2;:::given byan= ( 1)nforn IN has infinitely many terms even though theirvalues are repeated over and over. On the other hand, theset{( 1)n:n IN}is exactlythe set{ 1;1}consisting of two sequence (an)n=1;2;:::of real numbers is said toconvergeto the real numberaprovided that for each" >0 there exists a positive integerNsuch that|an a|< "whenevern > N. If (an)n=1;2;:::converges toa, we write limn!1an=a. The numberais called thelimitof the sequence (an)n=1;2.

2 A sequence that does not converge tosome real number is said thatlimn!11n= 0 given" >0, we wish to find a positive integerNsuch thatn > Nimplies|1n 0|< ". The latter is equivalent ton >1=". ChooseN= 1=" + 1. Ifn > N, thenn >1=", and hence|1n 0|< ". This shows that limn!11n= |r|<1, thenlimn!1rn= 0 0, thenrn= 0 for alln IN. Obviously, limn!1rn= 0 in this = 0. Since|r|<1, we have 1=|r|>1. Letb:= 1=|r| 1. Thenb >0 and1=|r|= 1 +b. It follows that|r|= 1=(1 +b) and|rn|= 1=(1 +b)n. By the Bernoulliinequality, (1 +b)n 1 +nbfor alln IN. Consequently,|r|n=1(1 +b)n 11 +nb<1nb:For given" >0, we wish to find a positive integerNsuch thatn > Nimplies|rn|< ".This happens if1nb< ", ,n >1=(b"). ChooseN:= 1=(b") + 1. Ifn > N, we haven >1=(b"), and hence|rn|< ". This shows limn!1rn= convergent sequence of real numbers has a unique (an)n=1;2;:::be a convergent sequence.

3 Suppose that limn!1an=sandlimn!1an=t. We wish to proves=t. For given" >0, by the definition of limit, thereexists a positive integerN1such thatn > N1implies|an s|< "=2:Moreover, there exists a positive integerN2such thatn > N2implies|an t|< "=2:Forn >max{N1; N2}, by the triangle inequality we have|s t|=|(s an) + (an t)| |an s|+|an t|<"2+"2=":This shows that|s t|< "for all" >0. It follows that|s t|= 0 and hences= sequence (an)n=1;2;:::of real numbers is said to beboundedif the set{an:n IN}is bounded in convergent sequence of real numbers is (an)n=1;2;:::be a convergent sequence such that limn!1an=a. For"= 1there exists a positive integerNsuch thatn > Nimplies|an a|<1:Forn > N, it follows that|an|=|a+ (an a)| |a|+|an a|<|a|+ 1:DefineM:= max{|a1|; : : : ;|aN|;|a|+ 1}. Then we have|an| Mfor alln IN. Hence,(an)n=1;2;:::is a bounded sequence (an)n=1;2;:::of real numbers is said todivergeto + provided that foreachM >0 there exists a positive integerNsuch thatan> Mwhenevern > N.

4 In thiscase we write limn!1an= + . Similarly, we say that (an)n=1;2;:::divergesto andwrite limn!1an= provided for eachM <0 there exists a positive integerNsuchthatan< Mwhenevern > is important to note that the symbols + and do not represent real limn!1an= + (or ), we shall say that the limit exists, but this does notmean that the sequence converges; in fact, it a sequence(an)n=1;2;:::of positive real numbers,limn!1an= + holds if and only iflimn!1(1=an) = limn!1an= + . Given" >0, letM:= 1=". Since limn!1an= + ,there exists a positive integerNsuch thatn > Nimpliesan> M= 1=". Consequently,n > Nimplies 1an 0 < ". This shows that limn!1(1=an) = limn!1(1=an) = 0. ForM >0, let":= 1=M. Since limn!1(1=an) = 0,there exists a positive integerNsuch thatn > Nimplies1an< ". Consequently,n > Nimpliesan>1="=M.

5 This shows that limn!1an= + .Example |r|>1, then the sequence (rn)n=1;2;:::is |r|>1, we have 1=|r|<1, and hencelimn!11|r|n= limn!1(1|r|)n= 0:By Theorem , it follows that limn!1|r|n= . Therefore, for|r|>1, the sequence(rn)n=1;2;:::is unbounded. 2. Limit Theorems for SequencesIn this section we will investigate some of the important properties of Sequences ofreal numbers. We start with algebraic operations on convergent !1an=aandlimn!1bn=b, thenlimn!1(an+bn) =a+bandlimn!1(an bn) =a given" >0, there exists a positive integerN1such thatn > N1implies|an a|< "=2:Moreover, there exists a positive integerN2such thatn > N2implies|bn b|< "=2:LetN:= max{N1; N2}. Ifn > N, then by the triangle inequality we have|(an bn) (a b)| |an a|+|bn b|<"2+"2=":This completes the (an)n=1;2;:::is a convergent sequence, then the above theorem tells us thatlimn!

6 1(an+1 an) = limn!1an+1 limn!1an= 0:Example := ( 1)nforn IN. The sequence (an)n=1;2; have|an+1 an|= 2 for alln IN. So the sequence (an)n=1;2; !1an=aandlimn!1bn=b, thenlimn!1(anbn) =ab:Moreover, ifbn = 0for alln INandb = 0, thenlimn!1anbn= have|anbn ab|=|anbn anb+anb ab| |anbn anb|+|anb ab|=|an| |bn b|+|b| |an a|:By Theorem , there exists a real numberM >0 such that|an| Mfor alln IN. ThenumberMcan be so chosen that|b| M. For given" >0, since limn!1an=aandlimn!1bn=b, there exists a positive integerNsuch thatn > Nimplies|an a|<"2 Mand|bn b|<"2M:Consequently, ifn > N, then|anbn ab| |an| |bn b|+|b| |an a| M|bn b|+M|an a|< ":This shows that limn!1(anbn) = handle quotients of Sequences , we first deal with reciprocals. We begin by consid-ering the equality 1bn 1b = b bnbnb =|b bn||bn| |b|:For given" >0, since limn!

7 1bn=b = 0, there exists a positive integerN1such thatn > N1implies|bn b|<|b|=2:It follows that|b|=|bn+(b bn)| |bn|+|b bn|<|bn|+|b|=2. Consequently, forn > N1we have|bn|>|b|=2 and 1bn 1b =|b bn||bn| |b| |b bn||b|2=2:There exists a positive integerN > N1such thatn > Nimplies|bn b|< "|b|2=2:Hence, forn > Nwe have 1bn 1b |b bn||b|2=2< ":4 This shows that limn!1(1=bn) = 1=b. Since limn!1an=a, by the first part of thetheorem we obtainlimn!1anbn= limn!1an 1bn= limn!1anlimn!11bn=a 1b=ab:This completes the limn!1an, wherean:=n3+ 6n2+ 74n3+ 3n 4; n IN:Solution. We havean=n3(1 +6n+7n3)n3(4 +3n2 4n3)=1 +6n+7n34 +3n2 4n3:Since limn!11n= 0, by Theorem we havelimn!11n2= limn!1(1n)(1n)= 0 and limn!11n3= limn!1(1n2)(1n)= 0:By Theorems and , it follows thatlimn!1(1 +6n+7n3)= 1 and limn!1(4 +3n2 4n3)= 4:Applying Theorem again, we obtainlimn!

8 1an=limn!1(1 +6n+7n3)limn!1(4 +3n2 4n3)=14:Theorem thatlimn!1an=aandlimn!1bn=b. If there is somen0 INsuch thatan bnfor alln n0, thena thata > b. Let":= (a b)=2>0. Since limn!1an=a, there exists apositive integerN1such thatn > N1impliesa " < an< a+":Since limn!1bn=b, there exists a positive integerN2such thatn > N2impliesb " < bn< b+":LetN:= max{N1; N2; n0}. Then forn > Nwe havebn< b+"=a " < an;which contradicts the assumption thatan bnfor alln n0. Thus we conclude thata := 0 andbn:= 1=nforn IN. Thenan< bnfor alln IN. Butlimn!1an= 0 = limn!1bn. So we do not have a strict inequality for the (an)n=1;2;:::,(bn)n=1;2;:::, and(xn)n=1;2;:::be three Sequences of realnumbers. Suppose thatlimn!1an= limn!1bn=s. If there exists somen0 INsuchthatan xn bnfor alln n0, thenlimn!1xn= " >0 be given. Since limn!1an=s, there exists a positive integerN1suchthatn > N1impliess " < an< s+":Since limn!

9 1bn=s, there exists a positive integerN2such thatn > N2impliess " < bn< s+":LetN:= max{N1; N2; n0}. Then forn > Nwe havean xn bnand hances " < xn< s+":This shows that limn!1xn= above theorem is often called thesqueeze theorem. The following two examplesillustrate applications of the limn!1|an|= 0, then limn!1an= that |an| an |an|. Since limn!1|an|= 0, we have limn!1 |an|= Theorem we conclude that limn!1an= 0. In particular, foran= ( 1)n=n, weobtain limn!1( 1)n=n= >0, then limn!1a1=n= , consider the casea 1. In this case, we have1 a1=n 1 +a 1n n IN:The first inequality is valid because 1n= 1 a. The second inequality comes from theBernoulli inequality. Indeed, since (a 1)=n 0, the Bernoulli inequality gives(1 +a 1n)n 1 +na 1n= 1 + (a 1) =a:Since limn!11 = 1 and limn!1(1 + (a 1)=n)= 1, by the squeeze theorem we obtainlimn!

10 1a1=n= remains to deal with the case 0< a <1. In this case, 1=a >1. By what has beenproved, limn!1(1=a)1=n= 1. Therefore, limn!1a1=n= limn!11=(1=a)1=n= generally, ifa >0 and if ( n)n=1;2;:::is a sequence of rational numbers such thatlimn!1 n= 0, then limn!1a n= 1. To prove this result, we first consider the casea >1. Since limk!1a1=k= 1 and limk!1a 1=k= 1, for any given" >0, there existssomeK IN such that1 " < a 1=K< a1=K<1 +":But limn!1 n= 0. Hence, there exists somen0 IN such that 1=K < n<1=Kwhenevern n0. Thus, forn n0we havea 1=K< a n< a1=K:Consequently, 1 " < a n<1 +"whenevern n0. This proves limn!1a n= 1. Theproof for the casea= 1 is trivial. It remains to consider the case 0< a <1. In this case,we have 1=a >1 and limn!1a n= limn!1(1=a) n= 1. This completes the (an)n=1;2;:::and(bn)n=1;2;:::be two Sequences of real numbers.